Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int r[105], c[105], tmp_r[105], tmp_c[105], g[105][105];
int n, m;
const int INF = 0x3f3f3f3f;
int solve(int up) {
  int flag = 0, ans = INF, tmp = 0;
  for (int i = 0; i <= up; i++) {
    flag = 0;
    tmp_r[0] = i;
    tmp = i;
    for (int j = 0; j < m; j++) {
      tmp_c[j] = g[0][j] - i;
      tmp += tmp_c[j];
    }
    for (int j = 1; j < n; j++) {
      tmp_r[j] = g[j][0] - tmp_c[0];
      tmp += tmp_r[j];
    }
    for (int j = 0; j < n; j++) {
      if (flag == -1) break;
      for (int k = 0; k < m; k++) {
        if (tmp_r[j] < 0 || tmp_c[k] < 0 || g[j][k] != tmp_r[j] + tmp_c[k]) {
          flag = -1;
          break;
        }
      }
    }
    if (flag != -1 && tmp < ans) {
      for (int j = 0; j < n; j++) {
        r[j] = tmp_r[j];
      }
      for (int j = 0; j < m; j++) {
        c[j] = tmp_c[j];
      }
      ans = tmp;
    }
  }
  return ans == INF ? -1 : ans;
}
void print() {
  for (int i = 0; i < n; i++) {
    while (r[i]) {
      printf("row %d\n", i + 1);
      r[i]--;
    }
  }
  for (int i = 0; i < m; i++) {
    while (c[i]) {
      printf("col %d\n", i + 1);
      c[i]--;
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &g[i][j]);
    }
  }
  int up = 505;
  for (int i = 0; i < m; i++) {
    up = min(g[0][i], up);
  }
  int ans = solve(up);
  cout << ans << endl;
  if (ans != -1) print();
  return 0;
}